If ${S^o}$ for ${H_2},C{l_2}$ and $HCl$ are 0.13, 0.22 and 0.19 $kJ{K^{ - 1}}mo{l^{ - 1}}$ respectively. The total change in standard entropy for the reaction
\[{H_2} + C{l_2} \to 2HCl\] is:
A) $30J{K^{ - 1}}mo{l^{ - 1}}$
B) $40J{K^{ - 1}}mo{l^{ - 1}}$
C) $60J{K^{ - 1}}mo{l^{ - 1}}$
D) $20J{K^{ - 1}}mo{l^{ - 1}}$
Answer
Verified
466.5k+ views
Hint: We are given the standard entropy values for ${H_2},C{l_2}$ and $HCl$. The total change in standard entropy for any reaction is the difference between standard entropy for products and standard entropy for reactants.
Complete step by step solution:
Let us first write the given values in the question.
Standard entropy for ${H_2}$ ($S_{{H_2}}^o$)= 0.13 $kJ{K^{ - 1}}mo{l^{ - 1}}$.
Standard entropy for $C{l_2}$ ($S_{C{l_2}}^o$) = 0.22 $kJ{K^{ - 1}}mo{l^{ - 1}}$.
Standard entropy for HCl ($S_{HCl}^o$) = 0.19 $kJ{K^{ - 1}}mo{l^{ - 1}}$.
Given reaction is: \[{H_2} + C{l_2} \to 2HCl\]
To find total change in standard entropy for the above reaction, we need to take the difference of the sum of standard entropy for products and sum of standard entropy for reactants.
Thus, we can also write, total change in standard entropy (\[\Delta {S^o}\]) as:\[\Delta {S^o} = \sum n{S^o}({\text{products}}) - \sum n{S^o}({\text{reactants}})\]
Here, n is the stoichiometric coefficient before products and reactants.
In the given reaction, the product is HCl and ${H_2}$and $C{l_2}$ are reactants.
Therefore, total change in standard entropy for the given reaction = $2(S_{HCl}^o) - (S_{{H_2}}^o + S_{C{l_2}}^o)$ . (Here, 2 is the stoichiometric coefficient of HCl).
Total change in standard entropy for the given reaction = $2(0.19) - (0.13 + 0.22) = 0.03KJ{K^{ - 1}}mo{l^{ - 1}} = 30J{K^{ - 1}}mo{l^{ - 1}}$ .
Thus, option A is the correct answer.
Note: It should be known that entropy change for a chemical reaction can be determined if standard or absolute entropies of each substance are known. The standard entropy of a substance is given by the symbol ${S^o}$and this is determined under standard conditions which are called STP conditions. The SI units for entropy are ${K^{ - 1}}mo{l^{ - 1}}$. As the temperature of the substance increases, its entropy increases because of an increase in the randomness of the molecules within the system.
Complete step by step solution:
Let us first write the given values in the question.
Standard entropy for ${H_2}$ ($S_{{H_2}}^o$)= 0.13 $kJ{K^{ - 1}}mo{l^{ - 1}}$.
Standard entropy for $C{l_2}$ ($S_{C{l_2}}^o$) = 0.22 $kJ{K^{ - 1}}mo{l^{ - 1}}$.
Standard entropy for HCl ($S_{HCl}^o$) = 0.19 $kJ{K^{ - 1}}mo{l^{ - 1}}$.
Given reaction is: \[{H_2} + C{l_2} \to 2HCl\]
To find total change in standard entropy for the above reaction, we need to take the difference of the sum of standard entropy for products and sum of standard entropy for reactants.
Thus, we can also write, total change in standard entropy (\[\Delta {S^o}\]) as:\[\Delta {S^o} = \sum n{S^o}({\text{products}}) - \sum n{S^o}({\text{reactants}})\]
Here, n is the stoichiometric coefficient before products and reactants.
In the given reaction, the product is HCl and ${H_2}$and $C{l_2}$ are reactants.
Therefore, total change in standard entropy for the given reaction = $2(S_{HCl}^o) - (S_{{H_2}}^o + S_{C{l_2}}^o)$ . (Here, 2 is the stoichiometric coefficient of HCl).
Total change in standard entropy for the given reaction = $2(0.19) - (0.13 + 0.22) = 0.03KJ{K^{ - 1}}mo{l^{ - 1}} = 30J{K^{ - 1}}mo{l^{ - 1}}$ .
Thus, option A is the correct answer.
Note: It should be known that entropy change for a chemical reaction can be determined if standard or absolute entropies of each substance are known. The standard entropy of a substance is given by the symbol ${S^o}$and this is determined under standard conditions which are called STP conditions. The SI units for entropy are ${K^{ - 1}}mo{l^{ - 1}}$. As the temperature of the substance increases, its entropy increases because of an increase in the randomness of the molecules within the system.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success
Master Class 11 Computer Science: Engaging Questions & Answers for Success
Master Class 11 Maths: Engaging Questions & Answers for Success
Master Class 11 Social Science: Engaging Questions & Answers for Success
Master Class 11 Economics: Engaging Questions & Answers for Success
Master Class 11 Business Studies: Engaging Questions & Answers for Success
Trending doubts
10 examples of friction in our daily life
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
State and prove Bernoullis theorem class 11 physics CBSE
Pigmented layer in the eye is called as a Cornea b class 11 biology CBSE
What problem did Carter face when he reached the mummy class 11 english CBSE